I.

\(2KNO3-->2KNO2+O2\)

\(n_{O2}=\frac{5,264}{22,4}=0,235\left(mol\right)\)

\(\)\(n_{KNO3}=2n_{O2}=0,47\left(mol\right)\)

\(m_{KNO3}=0,47.101=47,47\left(g\right)\)

II.

\(xAl+yO2-->2AlxOy\)

\(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\)

\(n_{AlxOy}=\frac{2}{x}n_{Al}=\frac{0,2}{x}\left(mol\right)\)

\(M_{AlxOy}=5,1:\frac{0,2}{x}=25,5x\)

\(x=2,y=3\Rightarrow CTHH:Al2O3\)

III. Cho 6,885g nhôm kim loại phản ứng với dung dịch axit H2S

O4 loãng có chứa 34,4g H2SO4 tinh khiết

a) Tính V khí H2 sinh ra (đktc)

b) Để thu đc lượng H2 trên thì phải dùng bn g kim loại sắ cho phản ứng với axit HCl dư

a) \(2Al+3H2SO4-->Al2\left(SO4\right)3+3H2\)

\(n_{Al}=\frac{6,885}{27}=0,255\left(mol\right)\)

\(n_{H2SO4}=\frac{34,4}{98}=0,35\left(mol\right)\)

\(\frac{0,255}{2}>\frac{0,35}{3}\Rightarrow Aldư\)

\(n_{H2}=n_{H2SO4}=0,35\left(mol\right)\)

\(V_{H2}=0,35.22,4=7,84\left(l\right)\)

b) \(Fe+2HCl--.>FeCl2+H2\)

\(n_{Fe}=n_{H2}=0,35\left(mol\right)\)

\(m_{Fe}=0,35.56=19,6\left(g\right)\)

Al₂O₃

4Al + 3O₂ --t^o--> 2Al₂O₃

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

___________0,15<------0,1

=> m_O2 = 0,15.32 = 4,8(g)

Bảo toàn KL: \(m_{Al}+m_{O_2}=m_{Al_2O_3}\)

\(\Rightarrow m_{O_2}=10,2-9=1,2(g)\)

n Al=\(\dfrac{32,4}{27}\)=1,2 mol

n O2=\(\dfrac{23,7984}{22,4}\)=1,062mol

4Al+3O₂-to>2Al₂O₃

1,2---------------0,6 mol

O2 dư

=>m _Al2O3=0,6.102=61,2g

2Al+6HCl->2AlCl₃+3H₂

1,2-----------------------1,8 mol

=>V_H2=1,8.22,4=40,32l

\(m_{Al} + m_{O_2} = m_{Al_2O_3}\)

Ta có : 

 \(n_{Al} = \dfrac{9}{27} = \dfrac{1}{3}(mol)\\ n_{Al_2O_3} = \dfrac{15}{102} = \dfrac{5}{34}(mol)\)

\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)

Theo PTHH : \(n_{Al\ pư} = 2n_{Al_2O_3} = \dfrac{5}{17} > n_{Al\ ban\ đầu}\)

Suy ra : Al dư.

Ta có :

 \(n_{O_2} = \dfrac{3}{2}n_{Al_2O_3} = \dfrac{15}{68}(mol)\\ \Rightarrow m_{O_2\ phản ứng} = \dfrac{15}{68}.32 = 7,059(gam)\)

số liệu không hợp lý

`a) PTHH:`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

` 0,2`                                        `0,3`      `(mol)`

`n_[Al] = [ 5,4 ] / 27 = 0,2 (mol)`

`b)V_[H_2] = 0,3 . 22,4 = 6,72 (l)`

`c)`

`AO + H_2` $\xrightarrow[]{t^o}$ `A + H_2 O`

`0,3`     `0,3`

`=> M_[AO] = 24 / [ 0,3 ] = 80 ( g // mol )`

`=> M_A = 80 - 16 = 64 ( g // mol )`

      `=> CTHH` của oxit đó là: `CuO`

a) \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

______0,3--------------->0,15

=> \(m_{Al_2O_3\left(PTHH\right)}=0,15.102=15,3\left(g\right)\)

=> m_{Al2O3 (thực tế)}  = \(\dfrac{15,3.100}{90}=17\left(g\right)\)

a) PTHH: Al + O₂ -> 2Al₂O₃

\(m_{Al_2O_3}=\dfrac{8,1.90\%}{100\%}=7,29\left(g\right)\)

\(\left(1\right).4Al+3O_2-^{t^o}\rightarrow2Al_2O_3\\ \left(2\right).m_{Al}+m_{O_2}=m_{Al_2O_3}\\ \left(3\right).m_{O_2}=m_{Al_2O_3}-m_{Al}=10,2-5,4=4,8\left(g\right)\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{50,4}{2.22,4}=0,45\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

LTL: \(\dfrac{0,2}{4}< \dfrac{0,45}{5}\rightarrow\) O₂ dư

Theo pthh: \(\left\{{}\begin{matrix}n_{Al_2O_3}=\dfrac{0,2}{2}=0,1\left(mol\right)\\n_{O_2\left(pư\right)}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(0,45-0,15\right).32=9,6\left(g\right)\\m_{Al_2O_3}=0,1.102=10,2\left(g\right)\end{matrix}\right.\)